OCR S1 2012 January — Question 1 4 marks

Exam BoardOCR
ModuleS1 (Statistics 1)
Year2012
SessionJanuary
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicDiscrete Probability Distributions
TypeOne unknown from sum constraint only
DifficultyEasy -1.3 This is a straightforward recall question requiring only that probabilities sum to 1 (giving a simple linear equation 0.1 + 0.3 + 2p + p = 1) and then computing E(X) using the standard formula. Both parts are direct application of basic definitions with no problem-solving or conceptual challenge.
Spec5.02b Expectation and variance: discrete random variables

1 The probability distribution of a random variable \(X\) is shown in the table.
\(x\)1234
\(\mathrm { P } ( X = x )\)0.10.3\(2 p\)\(p\)
  1. Find \(p\).
  2. Find \(\mathrm { E } ( X )\).

Question 1:
Part (i)
AnswerMarks Guidance
AnswerMarks Guidance
\(0.1 + 0.3 + 2p + p = 1\) oeM1
\(p = 0.2\)A1 [2]
Part (ii)
AnswerMarks Guidance
AnswerMarks Guidance
\(\Sigma xp\)M1 \(\geq 2\) terms correct, FT \(p\)
\(= 2.7\) oeA1f [2]
## Question 1:

### Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $0.1 + 0.3 + 2p + p = 1$ oe | M1 | |
| $p = 0.2$ | A1 [2] | |

### Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\Sigma xp$ | M1 | $\geq 2$ terms correct, FT $p$ | eg $\div 4$: M0A0 |
| $= 2.7$ oe | A1f [2] | |

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1 The probability distribution of a random variable $X$ is shown in the table.

\begin{center}
\begin{tabular}{ | c | c | c | c | c | }
\hline
$x$ & 1 & 2 & 3 & 4 \\
\hline
$\mathrm { P } ( X = x )$ & 0.1 & 0.3 & $2 p$ & $p$ \\
\hline
\end{tabular}
\end{center}

(i) Find $p$.\\
(ii) Find $\mathrm { E } ( X )$.

\hfill \mbox{\textit{OCR S1 2012 Q1 [4]}}